CHEMISTRY LEVEL 4C

(CHM415109)

INTRODUCTORY QUANTUM THEORY

& PERIODIC TABLE

THEORY SUMMARY

&

REVISION QUESTIONS

Tasmanian TCE Chemistry Revision Guides by Jak Denny are licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.

©JAK DENNY

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INDEX: PAGES

• INTRODUCTION 3

• DEVELOPMENT OF ATOMIC THEORY 4-5

• MASS SPECTROSCOPY 6-7

• IONIZATION ENERGY THEORY 8-9

• IONIZATION ENERGY QUESTIONS 10

• A QUANTUM MECHANICAL VIEW 11-12

• ELECTRON CONFIGURATIONS 13

• ORBITALS AND THEIR SHAPES 14

• ORBITAL ENERGY LEVEL DIAGRAM 15

• THE PERIODIC TABLE 16-17

• PROPERTIES OF PERIOD 3 ELEMENTS 18

• REVIEW QUESTIONS ON PERIOD 3 19

• GROUP IA (ALKALI METALS) 20-21

• GROUP VIIA (HALOGENS) 22-23

• REVIEW QUESTIONS 24-27

• REVISION TEST 28

• REVISION TEST ANSWERS 29

*CHECK: http://www.webelements.com/webelements/scholar/

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INTRODUCTION: The historical development of our understanding of matter extends back to the times of Democritus at around 400 B.C. when he proposed that all matter was comprised of particles that were not infinitely subdivisible. He proposed that the process of cutting a substance such as a piece of copper in half, again and again and again, would ultimately reach a point where one would have a particle that was the fundamental building block of copper. The word atom comes from the Greek ‘atomos’ which means particle. Other philosophers of the time such as Aristotle proposed that matter was infinitely subdivisible and no matter how many times you chopped the copper block in half, you would always have copper. The historical development of our present day understanding of chemistry hardly changed for the next 1800 years until chemists such as Dalton and Avogadro moved from the alchemists’ approach to a more scientific enquiry based study of matter. To attempt to show the steps in our developing knowledge of chemistry is difficult but it is possible to see 6 or 7 distinct phases of development where each led to a better understanding of the nature of atoms and atomic structure. i.e.

DATE PERSON CONCEPT 1803 John Dalton Atoms as particles 1904 J.J. Thomson Electrons and protons 1911 Ernest Rutherford Orbiting electrons 1913 Neils Bohr Electron energy levels 1932 James Chadwick Neutron discovered 1926 Schrodinger Quantum mechanical model 1932 Heisenberg Refinement of quantum model

HISTORICAL DEVELOPMENT OF ATOMIC STRUCTURE

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DEVELOPMENT OF THE ATOMIC THEORY:

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MASS SPECTROSCOPY: The first mass spectrometer was built by Francis Aston in 1919 and was used to confirm the existence of isotopes although the discovery of the neutron didn't occur until Chadwick's research in 1932. The basic principle involves combining electric and magnetic fields so that positive ions can be accelerated and then deflected by varying amounts in accordance with their mass. This is best understood if we use a more familiar model. (i) Steel ball bearings of different masses are rolled down a ramp with gravity being the accelerating field. (ii) At the bottom of the ramp, on one side, we place a magnet which attracts the steel ball bearings as they pass. (iii) The attracting effect of the magnet results in the ball bearings moving in circular paths where the radius is dependent upon the masses. (iv) The lightest ball bearing will be deflected most and the heaviest ball bearing is deflected least. (see diagram) This process of combining gravitational and magnetic fields to separate the ball bearings is essentially the basis of the mass spectrometer except that we are using an electric field to accelerate ions rather than a gravitational field to accelerate ball bearings! HOW DOES A MASS SPECTROMETER WORK? (i) The sample for analysis is firstly vaporised (if it is not already a gas). (ii) The gas then enters a low pressure ionisation chamber where it is bombarded with electrons resulting in the formation of mostly 1+ ions, i.e. A(g) + e− + energy → A+

(g) + 2e−

(iii) These ions are now accelerated to high speeds using an electric field. They move towards a negatively charged plate with a hole in its centre. (iv) A stream of positive ions passes through the hole and is then acted on by a magnetic field at right angles to the electric field. (v) This magnetic field causes the 1+ ions to move in curved paths with radius proportional to their masses. (vi) By changing the strengths of the electric and magnetic fields, different mass ions will arrive at the detector at different field strengths. (vii) A detector identifies the ions and their abundance. The detection is based upon either: - the current produced by the ions, or - the intensity of the 'spot' formed on a photographic plate. (viii) The picture/graph obtained is called a "mass spectrogram".

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SIMPLIFIED DIAGRAM OF A MASS SPECTROMETER: Q1. Magnesium exists as three naturally occurring stable isotopes; i.e. magnesium-24 (24Mg) natural abundance = 79% magnesium-25 (25Mg) natural abundance = 10% magnesium-26 (26Mg) natural abundance = 11% Draw the mass spectrum for Mg. Q2. Chlorine exists as two naturally occurring stable isotopes; i.e. 35Cl and 37Cl. When analysed in a mass spectrometer, the molecular ions formed are Cl2+

(g). How many peaks will appear in the mass spectrum and at what mass numbers will they appear? ANS. Three peaks will occur, one at each of mass numbers 70, 72 and 74. Q3. On the right is shown the mass spectrum for tungsten (W). (i) Use this mass spectrum to identify the isotopes of tungsten that occur naturally. (ii) Use the answer for part (i) and the relative abundances to determine the Ar(W). (183.7) (iii) Find the number of neutrons in 1 million tungsten atoms. (109.7 million)

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IONIZATION ENERGY: (i) DEFINITION: The ionization energy of an atom is the least amount of energy required to remove the most loosely bound electron from a gaseous atom. e.g. for magnesium the energy required is 733 kJ per mole of Mg atoms: Mg(g) + 733 kJ → Mg(g)

+ + e− The removal of electrons can be achieved by two different means of energy input using: (i) light energy ('photo-ionization') (ii) electrons ('electron bombardment') (ii) TRENDS IN THE FIRST IONIZATION ENERGY: The ionization energies (E1) for the first 20 elements are shown below. ATOMIC NO. ELEMENT E1 (kJ/mol) 1 H 1312

2 He 2371

3 Li 520

4 Be 899

5 B 800

6 C 1086

7 N 1403

8 O 1313

9 F 1680

10 Ne 2080

11 Na 495

12 Mg 733

13 Al 577

14 Si 786

15 P 1063

16 S 999

17 Cl 1256

18 Ar 1520

19 K 418

20 Ca 590

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When the ionization energy data from the previous page are plotted as shown below, we notice: (i) relative minima occur for Li, Na and K suggesting that these elements can lose an electron easily. (ii) relative maxima occur for He, Ne and Ar suggesting that these elements cannot lose an electron easily. (iii) First ionization energies show a general increase across each period due to the outer electrons being in the same valence orbitals (thus approx. the same distance from the nucleus) and yet attracted by an increasing nuclear charge.

ATOMIC NUMBER

(iii) SUCCESSIVE IONIZATION ENERGIES: Although we are often considering just the removal of the first electron (i.e. the first ionization energy "E1") chemists have learnt a good deal of information about electron energy levels by measuring successive ionization energies. The removal of a second electron requires the second ionization energy given by E2, the removal of a third electron requires the third ionization energy given by E3 .........etc e.g. Be(g) + 899 kJ → Be(g)

+ + e− (E1 = 899 kJ) Be(g)

+ + 1748 kJ → Be(g)

2+ + e− (E2 = 1748 kJ) Be(g)

2+ + 14780 kJ → Be(g)

3+ + e− (E3 = 14780 kJ) Be(g)

3+ + 20940 kJ → Be(g)

4+ + e− (E4 = 20940 kJ) There is no fifth ionization energy for beryllium (E5) because there are only 4 electrons per Be atom to be removed! By looking at the data above we can see that the first and second ionization energies for beryllium are both fairly low (both under 2000 kJ) whereas the third I.E. shows a huge increase to nearly 15000 kJ. This suggests that Be atoms have two outer or valence electrons that are easily removed.

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(iv) QUESTIONS ON IONIZATION ENERGY: Q1. Why do the first ionization energies of elements show a general: (i) increase across any given period of the Periodic Table. (ii) decrease down any given group of the Periodic Table. Q2. Why does boron have a lower ionization energy than beryllium even though boron has a higher nuclear charge? Q3. An element X has successive ionization energies of E1 = 733 kJ/mole, E2 = 1440 kJ/mole and E3 = 15780 kJ/mole respectively. (a) How many valence electrons does X possess? (b) Which group of the Periodic Table does X probably belong to? Q4. In terms of ionization energies and orbital occupancy explain why aluminium forms 3+ ions but does not form 4+ ions. Q5. As Na+ is isoelectronic (has the same number of electrons) with Ne, compare the first ionization energy of neon with the second ionization of sodium. Q6. (i) Why is the second ionization energy of an atom always greater than the first I.E.? (ii) Why does this rule in part (i) not apply to the element hydrogen? Q7. The ionization of both atomic hydrogen and helium involves the removal of a 1s electron and yet E1 for He is much greater than E1 for hydrogen. Explain. Q8. Qualitatively sketch a graph of the first 5 successive I.E.s for: (i) sodium (ii) silicon (iii) aluminium (iv) calcium. Q9. Given that the first ionization energy for magnesium is 733 kJ/mol, would the value of E1 for an excited Mg atom be greater or less than 733 kJ/mol? Explain. Q10. Element X is investigated and found to have: First ionization energy (E1) = 1,290 kJ mol-1. Second ionization energy (E2) = 13,680 kJ mol-1. Third ionization energy (E3) = 15,070 kJ mol-1.

(i) To which group of the Periodic Table does X most likely belong? Explain your reasoning.

(ii) What is the likely formula for: • the oxide of element X? • the chloride of element X?

(iii) Why is the third ionization energy (E3) greater than the second ionization energy (E2)?

Q11. Explain why the decreasing ionization energies as we progress down Group I of the periodic table mean that the elements become increasingly strong reducing agents whereas the decreasing ionization energies as we progress down Group VII means that the elements become increasingly weak oxidisers.

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ATOMIC STRUCTURE A QUANTUM MECHANICAL VIEW 1. EMISSION SPECTRA (a) Emission of light from a heated or bombarded atom results from a sequence of steps that make up an "ELECTRONIC TRANSITION". In summary: - all atoms are normally in the lowest energy or "GROUND" state. - absorption of energy occurs by heating or e- bombardment - electron(s) are "PROMOTED" to higher energy levels - the atom is now said to be in an "EXCITED" state: this occurs only momentarily - electron(s) now drop back through an energy gap ∆E to the lower energy levels - energy is now "EMITTED" in the form of a photon (quantum) of light - the photon's energy (colour/type) is determined by the energy gap ∆E - frequency of the emitted light (f) is given by the Planck Equation ∆E = hf where h = Planck's Constant = 3.99 x 10−13 kJ s mol−1 - the atom is now back in its "ground" state. (b) The emission spectra for the various elements are like 'fingerprints', uniquely identifying each element. (c) Because only a limited number of energy lines are found in each emission spectrum, we conclude that there must be a limited number of possible energy states for an atom; ie atoms are "QUANTIZED" (d) The "bookshelf" analogy is a useful way of explaining this concept; i.e. books can only be on shelves 1,2,3, ............. but not in-between! WAVE MODEL FOR LIGHT & ELECTRONS: (a) In 1923 de Broglie postulated the duality of light - suggesting that light behaved as both particles (photons/quanta) and waves. (b) By 1926 Schrödinger had extended the particle~wave model to explain the properties of electrons. (c) This new approach was quite different from the classical Newtonian physics and was named QUANTUM MECHANICS. (d) A key feature of the quantum mechanical view of the atom is that electrons are located in regions of the atom called ORBITALS which are cloud-like regions surrounding the nucleus. The energy of each orbital is found by solving complex quantum mechanical equations. (e) An orbital can be thought of as a region in which the probability of locating the electron is greater than a certain value e.g. 95% (c.f. bees around a hive).

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SHELLS, SUBSHELLS & ORBITALS: (i) Shells are the major energy levels (historically called K, L, M, N,... ) (ii) Within each shell there are subshells of similar but not identical energy (iii) These subshells are labelled s, p, d, f,....... (iv) 's' subshells are spherical regions (clouds) and 'p' are dumbell shaped etc. (v) subshells within a shell are made up of orbitals all of which have identical energy

‘s’subshells have only 1 orbital ‘p’ subshells have 3 orbitals 'd' subshells have 5 orbitals 'f' subshells have 7 orbitals ‘g’ subshells have 9 orbitals etc…………….

ELECTRON ACCOMMODATION IN ORBITALS: (i) Pauli Exclusion Principle: "An atomic orbital can hold a maximum of 2 electrons, i.e. it may hold 0, 1 or 2 e-" (ii) The two electrons that occupy the same orbital have the same energy but have opposite angular momentum i.e. opposite "spin". (iii) Hund's Rule: "All orbitals of equal energy acquire 1 electron before any accepts 2" (iv) Electrons occupy the lowest possible energy levels except when excitation occurs. ie. atoms are normally in the "ground state". (v) The order of filling of the orbitals in increasing energy level in a many electron atom is: i.e. the filling sequence is: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4 f< 5d < ...........

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EXAMPLES OF ELECTRON CONFIGURATIONS: Using the filling sequence for the orbitals as discussed above, we see that the ground state electronic configuration for any atom or ion is simply found by allocating the number of electrons to the lowest possible vacant orbitals. For example: (a) Hydrogen: H = 1s1 (b) Sodium: Na = 1s2 2s2 2p6 3s1 or (Ne) 3s1 (c) Oxygen: O = 1s2 2s2 2p4 or (He) 2s2 2p4

(d) Zinc: Zn = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 or (Ar)4s2 3d10 etc………..

Q12. Give the ground state configuration for: (i) a potassium atom (ii) a sulfur atom (iii) a calcium atom (iv) an aluminium atom (v) a chlorine atom (vi) an aluminium ion (vii) a fluoride ion (viii) a rubidium ion Q13. The ground state electronic configuration for a sulfide ion (S2−) is the same as the ground state electronic configuration for an argon atom. Based on this fact: (i) what are the similarities between S2− and Ar? (ii) what are some of the differences between S2− and Ar? (iii) compare the size (radii) of S2− and Ar. Q14. Give the ground state electronic configuration for each of the EIGHT elements that comprise PERIOD 3. Q15. An electrically neutral atom has the following electronic configuration: X = 1s2 2s2

2p6 3s2

3p6 4s1

3d2

(i) identify the element X (ii) what is special about this atom of element X? Q16. Give the general form for the electronic configuration for: (i) the “alkali metals” i.e. Group IA (ii) the “alkali earths” i.e. Group IIA Q17. An element has as part of its valence orbital electron configuration the following arrangement: X = ( )…...ns2

np5 where ( ) = noble gas

(i) what group of the Periodic Table does X belong to? (ii) how many valence electrons does X have. (iii) what is the formula of the compound formed between X and calcium? (iv) draw the electron-dot representation for the compound formed by X and sulfur. Q18. Give the name, symbol and ground state electronic configuration for the element that is in: (i) period 2 , group V which is also called period 2 , group 15 (ii) period 3 , group IV which is also called period 3 , group 14 (iii) period 4 group III which is also called period 4, group 13 (iv) period 3 group IV which is also called period 3 , group 14

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ORBITALS & THEIR SHAPES: Remember that the words orbital and orbit are very different in their meanings. An orbit involves a very predictable path being followed such as occurs with the Moon’s movement around the Earth. The term ‘orbital’ refers to a region of space surrounding an atom where an electron is likely to be located within a given probability. Just as we can specify the likelihood of locating bees around a hive to a given probability, so too can we specify the probability of locating electrons. Rather than being distinct pathways as occur with orbits, orbitals are thus probability distributions are appear as three dimensional ‘cloud’ like regions with cloud-like shapes depending upon the orbitals concerned. ‘s’ orbitals are spherical in shape. ‘p’ orbitals are dumbbell in shape (somewhat ‘propeller’ shaped). ‘d’ orbitals are double dumbbell in shape or even more complex.

SHAPES OF ORBITALS

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A SCHEMATIC ENERGY LEVEL DIAGRAM OF THE ORBITALS FOR A MANY ELECTRON ATOM:

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THE PERODIC TABLE The information gained from ionization energies enables us to come up with the model for electron arrangement as shown on page 15. Long before such a model had been developed, chemists recognised similarities between certain elements and established so-called “Chemical Families” which we now associate with the various ‘groups’ within the Periodic Table. GROUPS OF THE PERIODIC TABLE Elements within any given vertical column (group) of the Periodic Table show chemical and physical similarities due to them having similar outer shell electron arrangements. EXAMPLE 1. All the elements in Group IA (also written as Group 1) have a ground-state electronic configuration of:

M = (noble gas) ns1 where M = Li, Na, K, Rb, Cs, Fr,…….…..

This gives a family of elements that we call the ALKALI METALS. EXAMPLE 2. All the elements in Group IIA (also written as Group 2) have a ground-state electronic configuration of:

M’ = (noble gas) ns2 where M’ = Be, Mg, Ca, Sr, Ba, Ra,…….…..

This gives a family of elements that we call the ALKALI EARTHS. EXAMPLE 3. All the elements in Group VIIA (also written as Group 17) have a ground-state electronic configuration of:

X = (noble gas)…..ns2 np5 where X = F, Cl, Br, I, At,…….…..

This gives a family of elements that we call the HALOGENS. EXAMPLE 4. All the elements in Group VIIIA (also written as Group 18) have a ground-state electronic configuration of:

X’ = (noble gas)…..ns2 np6 where X = Ne, Ar, Kr, Xe, Rn,…….

This gives a family of elements that we call the NOBLE GASES.

~~~~~~~~~~~~~~~~~~~~~

Q19. Give the ground state electronic configuration for each of the Alkali Metals and explain why these elements all have the tendency to form 1+ ions but never 2+ ions. Q20. What would be the general format for the ground state electronic configuration for the elements that comprise Group VIA (Group 16) which are sometimes referred to as the CHALCOGENS? (noble gas)…..ns2 np4 Q21. In example 4 above, why was the element helium (He) a special case and not represented by the standard format X' = ( )…..ns2 np6 even though it is a noble gas?

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PERIODS OF THE PERIODIC TABLE Elements across any given horizontal row of the Periodic Table make up a PERIOD of the Periodic Table. Each period corresponds to the filling of a particular main quantum level and because these levels can accommodate differing numbers of electrons, the periods differ in length. For example:

PERIOD

ELEMENTS

NUMBER OF ELEMENTS

1

H ↔ He

2

2

Li ↔ Ne

8

3

Na ↔ Ar

8

4

K ↔ Kr

18

5

Rb ↔ Xe

18

6

Cs ↔ Rn

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Elements within the same period do not show chemical and physical similarities but show a trend of changing properties as electrons are progressively added to the same quantum level. e.g. PERIOD 2 starts with Li = (He) 2s1 and the second level fills up and ends with Ne = (He) 2s2 2p6

e.g. PERIOD 3 starts with Na = (Ne) 3s1 and the third level fills until we end with Ar = (Ne) 3s2 3p6

Across any given period of the Periodic Table, the general trends in properties are typically: LEFT SIDE OF PERIODIC TABLE ↔ RIGHT SIDE OF PERIODIC TABLE Group IA metallic elements ↔ Group VIIIA non-metallic elements

Group IA basic oxides ↔ Group VIIA acidic oxides

Group IA electropositive elements ↔ Group VIIA electronegative elements

Group IA powerful reducing agents ↔ Group VIIA powerful oxidisers

Group IA low ionization energy ↔ Group VIIIA high ionization energy

A more detailed view of the changing properties across the Periodic Table is shown on the next page (18) where the variations across period 3 are shown.

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REVIEW QUESTIONS ON PERIOD 3 ELEMENTS Q22. Why do the first ionization energies (E1) of the individual elements increase across a given period such as period 3 of the Periodic Table? Q23. Consider the 8 elements that comprise period 3; i.e. Na Mg Al Si P S Cl Ar ANS. (i) Which element has the lowest first ionization energy (E1)? (Na) (ii) Which element has the highest first ionization energy (E1)? (Ar) (iii) Which element has the highest second ionization energy (E2)? (Na) (iv) Which element does not react to form any compounds? (Ar) (v) Which element is the most reactive non-metal? (Cl) (vi) Which element is the most reactive metal? (Na) (vii) Which elements form basic oxides? (Na, Mg) (viii) Which of these elements form acidic oxides? (P, S, Cl) (ix) Which element is a semi-conductor of electricity? (Si) (x) Which element exists as a covalent network solid? (Si) (xi) Which element forms an ionic chloride with formula XCl2? (Mg) (xii) Which element forms a covalent chloride with formula XCl2? (S) (xiii) Which element has the lowest boiling point? (Ar) (xiv) Which element forms acidic oxides with the formulae XO2 and XO3? (S) (xv) Which element forms an amphoteric oxide? (Al) (xvi) Which element exists as tetra-atomic molecules X4? (P) (xvii) Which elements exist in two or more allotropic forms? (P, S) (xviii) Which element forms a covalent chloride XCl3? (P) (xix) Which element exists as diatomic molecules X2? (Cl)

(xx) Which element has the greatest difference between the third and fourth ionization energies; i.e. E3 and E4? (Al)

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GROUP IA (ALKALI METALS) PROPERTIES: ELEMENT

ELECTRON CONFIGURN

E1 (kJ mo1-1)

ATOMIC RADIUS (pm)

M.P. (oC)

DENSITY (g mL−1)

Li (He) 2s1 526 67 180 0.53 Na (Ne) 3s1 504 98 98 0.97 K (Ar) 4s1 425 135 64 0.86

Rb (Kr) 5s1 410 148 39 1.53 Cs (Xe) 6s1 380 167 29 1.87

(E1 = first ionization energy, pm = picometres = 10−12m and M.P = melting point) (i) This chemical family comprises the highly reactive metallic elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs) and francium (Fr). Francium is a synthetic and radioactive element that will not be of great concern to us in this discussion. (ii) All these metallic elements have just one valence electron; i.e. one more electron than the nearest noble gas. (iii) They all have very low first ionization energy (E1) values and very high second ionization energy (E2) values. The values of E1 decrease down the group. (iv) These metals react readily by the loss of one electron thus forming 1+ ions where the 1+ ion achieves stability by having a noble gas electron configuration. This high tendency to form positive ions is why the alkali metals are described as electropositive. e.g. Na → Na+ + e− where Na+ = (Ne) (v) As highly reactive metals that are electron donors, these elements are powerful reducing agents. The alkali metal reactivity increases down the group because the outer single valence electron is progressively further and further from the nucleus and thus not as strongly held and therefore more easily lost.

(vi) Each of the alkali metals reacts explosively with water to form the metal hydroxide and hydrogen gas. The formation of hydroxyl ions makes the solution alkaline or basic. e.g. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) NaOH(aq) → Na+

(aq) + OH–(aq)

(Note that by knowing any chemical equations for the reaction of sodium; eg with water as shown, one can immediately predict what the equations will be for any member of this family. One only has to replace sodium’s symbol in the equation with the other alkali metal’s symbol e.g., rubidium reacting with water gives ……

2Rb(s) + 2H2O(l) → 2RbOH(aq) + H2(g)

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GROUP IA (ALKALI METALS) CONTINUED: (vii) The reactivity of these metals increases with atomic number (i.e. down the group) due to the outer valence electron being further from the nucleus and thus, less strongly attracted; (see diagram below) (viii) The alkali metals react rapidly with oxygen at room temperature to form metal oxides and even metal peroxides e.g. 4Na(s) + O2(g) → 2Na2O(s) (product is sodium oxide) 2Na(s) + O2(g) → Na2O2(s) (product is sodium peroxide) To prevent these reactions with O2 gas, the very reactive alkali metals are normally stored under paraffin oil or kerosene. (ix) All the commonly encountered compounds formed by the alkali metals are ionic with the alkali metal present in the M+ ionic state. Practically all of these ionic solids are white crystalline, water soluble compounds with high melting points.

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GROUP VII or 17 (THE HALOGENS) PROPERTIES: ELEMENT

ATOMIC NUMBER

ELECTRON CONFIGURN

ATOMIC RADIUS (pm)

M.P. (oC)

B.P. (oC)

F 9 (He) 2s2 2p5 68 −220 −188 Cl 17 (Ne) 3s2 3p5 99 −101 −35 Br 35 (Ar)..…4s2 4p5 114 −7 58 I 53 (Kr)..…5s2 5p5 133 114 183

At 85 (Xe)..…6s2 6p5 - - - (pm = picometres = 10−12m M.P = melting point B.P = boiling point) (i) These elements form a family with many similarities in terms of their properties although astatine is of limited concern to us here as it is highly radioactive and thus unstable. (ii) Each of the halogens is one electron less than a noble gas configuration and thus they tend to react by the gaining of one electron per atom and the subsequent formation of 1− halide ions.

e.g. Cl + e− → Cl− where Cl− = (Ar) The chloride ion has the same electron configuration as an argon atom. (iii) The halogens are powerful electron acceptors i.e. powerful oxidising agents. The oxidising strength decreases down the group because, as the atomic radius increases, the electron attracting capacity decreases; (see diagram next page) (iv) The high tendency to attract electrons and form negative ions means that the halogens are described as being electronegative elements. (v) As uncombined elements, the halogens exist as diatomic molecules X2 with an intramolecular single covalent bond and van der Waal’s forces between molecules; i.e. intermolecular.

: X : X : (vi) The physical properties of the halogen elements are determined by the van der Waal’s forces increasing with size and mass of the molecules. e.g.

HALOGEN FORMULA STATE AT S.L.C. COLOUR fluorine F2 gas pale yellow chlorine Cl2 gas pale green/yellow bromine Br2 liquid reddish brown iodine I2 solid deep purple

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GROUP VII or 17 (THE HALOGENS) CONTINUED: (vii) The halogens react with most metals to form ionic metal halides with the reactivity decreasing down the group; e.g. 2Al(s) + 3Br2(l) → 2AlBr3(s) (viii) The halogens react with many non-metals to form covalent halides, once again with the reactivity decreasing down the group; Examples: CBr4 PCl5 HF HBrO3 SF6 CI4

(ix) The hydrogen halides (HX) are water soluble covalent gases which, in aqueous solution ionise to give acidic solutions; i.e. HCl(aq) → H+

(aq) + Cl−(aq)

(x) The oxides of the halogens (e.g. Cl2O) also give acidic aqueous solutions. Cl2O(g) + H2O(l) → 2HOCl(aq) HOCl(aq) → H+

(aq) + OCl−(aq)

(xi) In the diagrams below, is used to represent the electron vacancy in the outer (valence) ‘p’ orbital of the halogen atom. The smaller the atomic radius, the greater the electron attracting capacity of the halogen; i.e. fluorine with the smallest radius is the most reactive element in this chemical family and the reactivity decreases down the group!

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PERIODIC TABLE REVIEW QUESTIONS Q24. Give the ground state electronic configuration for each of the EIGHT elements that comprise PERIOD 3. Q25. An electrically neutral atom has the following electronic configuration: X = 1s2 2s2

2p6 3s2

3p6 4s1

3d2

(i) Identify the element X (ii) What is special about this atom of element X? Q26. Give the general form for the electronic configuration for: (i) the "alkali metals" i.e. Group I (ii) the "alkali earths" i.e. Group II Q27. An element has as part of its valence orbital electron configuration the following arrangement: X = ( )...ns2

np5 where ( ) = noble gas

(i) what group of the Periodic Table does X belong to? (ii) how many valence electrons does X have. (iii) what is the formula of the compound formed between X and calcium?

(iv) draw the electron-dot representation for the compound formed by X and sulfur.

Q28. Give the name, symbol and ground state electronic configuration for the element that is in: (i) period 2 , group V (ii) period 3 , group IV (iii) period 4 group III Q29. Which element has the ground state electronic configuration of: (Xe) 6s2 4f14

5d10

Q30. Give the ground state electronic configuration for: (a) Kr (b) Br− (c) Rb+ - what do all three have in common? - what are three differences? - what is the order for the particle's atomic/ionic radii (smallest first) ? Explain. Q31. Why do the atomic radii of elements tend to DECREASE across any given period of the Periodic Table? Q32. Describe the trend in change of chemical properties as one progresses across any given row (period) of the Periodic Table. Q33. Why do the Alkali Metals become increasingly reactive down the group whereas the opposite is true for the Halogens?

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Q34. Consider the following elements with ground states shown: A = ( ) ns2

B = ( ) ns2

np5

C = ( ) ns2 np4 D = ( ) ns1

E = ( ) ns2 np6 F = ( ) ns2

np2

What is the most likely chemical formula for the compound formed (if any) between:

(i) A and C (ii) D and C (iii) B and F (iv) C and E (v) D and B (vi) C and F ? Q35. Give a graphical sketch of the successive ionization energies (E1 to E5) for the element magnesium. Q36. Use your graph from Q35 to explain why Mg forms 2+ ions but not 3+ ions in normal chemical reactions. Q37. Describe the steps involved in the process in which (e.g.) copper compounds when heated by a bunsen burner produce a bright green light. Q38. What are some of the predicted properties of the following unknown elements? (i) element 117 (ii) element 118 (iii) element 119 Q39. Before the year 1604, only 12 elements had been discovered. What scientific reasons account for the fact that about 100 more elements were later discovered? Q40. Why were the “Noble Gases” (Group VIII) amongst the last of the elements to be discovered? Q41. What is meant by the term “periodicity”? What are some examples of the properties of elements that vary periodically? Q42. Why did Mendeleev’s Periodic table have: (i) Zn in the same group as Ca? (ii) Ag in the same group as Na? Q43. Mendeleev based his Periodic Table on his calculated relative atomic masses (Ar). This would have meant that he initially placed iodine (mass = 126.9) in Group VI and the element tellurium (mass = 127.9) in Group VII. He realised this order was wrong but couldn’t explain the reason. What is the modern day explanation for having their order reversed? Q44. What was Ernest Rutherford’s nuclear model of the atom and what dramatic experiment led him to this nuclear model? Q45. Henry Moseley made a major discovery using X-rays which led to a new ordering concept for the elements. What was this development? Q46. The development of the Mass Spectrometer enabled chemists to identify different types of the same element. How was this achieved and what ‘new’ particle was postulated to explain this occurrence?

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Q47. Why was the neutron difficult to detect? Q48. What is the difference between an emission spectrum and an absorption spectrum? Give examples. Q49. The successive ionisation energies for any given element give a very important and direct indication of the energy levels for the electrons of that element. Sketch a graph of the likely successive ionisation energies for the element aluminium from E1 to E13. Q50. Describe at a basic level the quantum mechanical model of electron structure including the distinction between shells, subshells and orbitals. Q51. Selenium (Se) is immediately below sulfur (S) in group VI of the periodic table. What is the likely chemical formula for: (i) calcium selenide? (ii) sodium selenide Q52. An element has an atomic number of 27. Give the ground state electronic configuration for a neutral atom of this element. Q53. Which of the following electronic configurations does NOT represent an atom in its lowest energy (ground) state?

A. 1s2 2s2 2p5 B. 1s2 2s2 2p6 3s2 3p1 C. 1s2 2s2 2p6 3s2 3p5 4s1 D. 1s2 2s2 2p6 3s2 3p6 3d5 4s2

The next 2 questions refer to the elements with outer-shell configurations shown: "A" = (noble gas) ns2 "C" = (noble gas) ns2 np5

"B" = (noble gas) ns2 np4 "D" = (noble gas) ns1 Q54. Give the chemical formula and describe the likely bonding for the compound formed by (i) A and B. (ii) A and C. (iii) B and C. Q55. Which elements from period 3 would correspond to A, B, C and D? Q56. The term ‘isoelectronic’ means ‘possessing the same number of electrons’. Give the symbols and charges of: (i) three non-metallic ions that are isoelectronic with an argon atom. (ii) three metallic ions that are isoelectronic with an argon atom.

(iii) Arrange the 6 ions listed above in order of increasing ionic radius. Explain your reasoning.

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Q57. An outline of the Periodic Table is given below with only hydrogen and helium having their normal symbols.

(a) Using only the symbols given above, write the chemical formula for the likely compound formed (if any) between the following pairs of elements: (i) A and F (ii) B and G (iii) C and F (iv) D and G (v) A and G (vi) F and G (b) In the compound formed between G and hydrogen, what is the ox(H)? (c) In the compound formed between A and hydrogen, what is the ox(H)? (d) Using only the symbols given above, write the formula for a compound formed by E that would be mostly: (i) ionic (ii) covalent (e) Draw the electron dot representation for the molecules (i) F2 (ii) G2

Q58. The element hydrogen is sometimes shown in the Periodic Table as a group I element and sometimes as a group VII element. (i) What is one chemical property of hydrogen suggesting that it should be in group I? (ii) What is one chemical property of hydrogen suggesting that it should be in group VII? (iii) Why does it seem preferable that most modern Periodic Tables now show the element hydrogen as a ‘family’ on its own? (i.e. not in either group I or VII) Q59. Successive ionization energy data (E1 ↔ E5) are given below for 4 unknown elements A, B, C and D. The energy units are in kJ mol−1. ELEMENT E1 E2 E3 E4 E5

A 135 329 517 3166 3401 B 102 829 936 1158 1308 C 165 347 1370 1562 1705 D 115 902 1105 1326 1569

(a) Write down the equation showing the process occurring when: (i) element A undergoes its first ionization. (ii) element B undergoes its third ionization. (ii) element D undergoes its fifth ionization. (b) Which of the listed elements A, B, C or D is likely to be in group III of the Periodic Table? (c) Which two of the listed elements A, B, C or D are likely to be in the same group of the Periodic Table? Q60. A positive ion has a ground state electron configuration of 1s2 2s2 2p6. What are the likely possibilities for its identity?

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CHEMISTRY(LEVEL 4C) PERIODIC TABLE (CRITERION 7) TEST 2 (25 marks) Q1. In terms of orbital occupancy, write down the electronic configuration for: Ca............... Mn............... Hf............... Se...............

(4 marks) Q2. Which one of the three chemical species, Ar Cl− or K+ would have the greatest radius? Explain by referring to their electronic configurations. (4 marks) Q3. An outline of the Periodic Table is given below with only hydrogen and helium having their normal symbols.

(i) Using only the symbols given above, write the chemical formula for the likely compound formed between the following pairs of elements. (2 marks)

A and F................... B and G...................... C and F...................... D and G...................

(ii) How many valence electrons does element E possess? ......................................................................................................... (1 mark)

(iii) Using only the symbols given above write a formula for a compound formed by E that would be mostly: (a) covalent ............................... (b) ionic ......................... (1 mark)

Q4. A student states that Mg3+ does not exist in nature because Mg3+ does not involve a "noble-gas configuration". Explain this answer in more detail using an electron energy level diagram for magnesium. (4 marks) Q5. "The addition of an electron to a chlorine atom gives it the same electron configuration as argon, thus the chloride ion should be as unreactive as argon." Why is this conclusion incorrect? Explain briefly. (2 marks) Q6. Consider the 8 elements that comprise Period 3 (ie. Na ↔ Ar). (i) Give the formula for the chloride and oxide of each element (where they exist). (3 marks) (ii) In terms of trends across the table, what changes occur in: (a) the atomic radii of the elements? (b) the metallic nature of the elements? (c) the ionic nature of the chlorides? (d) the acid/base nature of the oxides? (4 marks)

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